Leibniz Formula, proof of alternating property

Question

$$F_{A} := \sum_{\sigma\in S_n}\operatorname{sign}(\sigma) \prod_{i=1}^n A_{i \sigma(i)}$$

I am trying the prove that $\det(A)=F(A)$. I know that to do this, I need to show that $F$ satisfies the multilinear, alternating, and normalized properties of the determinant. I have proven multilinear and normalized, but I do not know how to prove the alternating property.

ie. prove that

$$ F(v_1, \ldots, v_i, \ldots, v_j, \ldots, v_n) = -F(v_1, \ldots, v_j, \ldots, v_i, \ldots, v_n). $$

help?

Answer 1

$\newcommand{\sign}{\operatorname{sign}}$Let $\tau$ be the element of $S_n$ that switches $i$ and $j$. Then $$ \begin{align} F(v_1,\ldots,v_j,\ldots,v_i,\ldots,v_n) & = \sum_{\sigma \in S_n} \sign(\sigma)\prod_{i=1}^n A_{i\sigma(\tau(i))} \\ & =\sum_{\sigma \in S_n} \sign(\sigma \tau) \prod_{i=1}^nA_{i\sigma(i)} \\ & = \sum_{\sigma \in S_n} -\sign(\sigma) \prod_{i=1}^nA_{i\sigma(i)} \\ & = -F(v_1,\ldots,v_i,\ldots,v_j,\ldots,v_n) \end{align} $$

Answer 2

In full techni-gory detail:

Let $A'$ be $A$ with the $a,b$ columns swapped.

$(a \ b)$ denotes the permutation that switches $a,b$.

\begin{eqnarray} \det A' &=& \sum_\sigma \operatorname{sgn} \sigma \prod_i [A']_{i \sigma(i)} \\ &=& \sum_\sigma \operatorname{sgn} \sigma \prod_i [A']_{\sigma^{-1}(i) i} \\ &=& \sum_\sigma \operatorname{sgn} \sigma \left(\prod_{i \notin \{a,b\}} [A']_{\sigma^{-1}(i) i} \right) [A']_{\sigma^{-1}(a) a} [A']_{\sigma^{-1}(b) b} \\ &=& \sum_\sigma \operatorname{sgn} \sigma \left(\prod_{i \notin \{a,b\}} [A]_{\sigma^{-1}(i) i} \right) [A]_{\sigma^{-1}(b) a} [A]_{\sigma^{-1}(a) b} \\ &=& \sum_\sigma \operatorname{sgn} \sigma \left(\prod_{i \notin \{a,b\}} [A]_{((a\ b) \circ\sigma)^{-1}(i) i} \right) [A]_{((a\ b) \circ\sigma)^{-1}(a) a} [A]_{((a\ b) \circ\sigma)^{-1}(b) b} \\ &=& \sum_\sigma \operatorname{sgn} \sigma \prod_i [A]_{((a\ b) \circ\sigma)^{-1}(i) i} \\ &=& \sum_\sigma (-1)\operatorname{sgn}(((a\ b) \circ\sigma)) \prod_i [A]_{((a\ b) \circ\sigma)^{-1}(i) i} \\ &=& -\sum_\sigma \operatorname{sgn} \sigma \prod_i [A]_{\sigma^{-1}(i) i} \\ &=& -\sum_\sigma \operatorname{sgn} \sigma \prod_i [A]_{i \sigma(i)} \\ &=& - \det A \end{eqnarray}